Answer
The solution of the the given boundary-value problem is
$$
y(x)=(\frac{-1}{e^6-1})+(\frac{e^6}{e^6-1})e^{-6x}.
$$
Work Step by Step
$$
y^{\prime \prime}+6 y^{\prime }=0, \quad y(0)=1, \quad y(1)=0
$$
The given equation is a homogeneous linear equation. So the characteristic equation of the differential equation is
$$
r^{2}+6r =r(r+6)=0
$$
whose roots are
$$
r=0, -6
$$
Therefore, the general solution of the given differential equation is
$$
y(x)=c_{1} +c_{2}e^{-6x},
$$
Then
$$
1=y(0)=c_{1} +c_{2}e^{0} \Rightarrow c_{1} +c_{2}=1
$$
and
$$
0=y(1)=c_{1} +c_{2}e^{-6} \Rightarrow c_{1} +c_{2}e^{-6}=0
$$
so
$$
c_{2}=\frac{e^6}{e^6-1},\:c_{1}=1-\frac{e^6}{e^6-1}=\frac{-1}{e^6-1}
$$
Thus, the solution of the boundary-value problem is
$$
y(x)=(\frac{-1}{e^6-1})+(\frac{e^6}{e^6-1})e^{-6x}.
$$